Thread: De.

**Bert** · June 25th 2006, 10:35 AM

Hello,

Given is $y^{4}+4y=0$ Who can I solove this? Who can do this step by step?

Greets Thanks .

**Jameson** · June 25th 2006, 01:09 PM

By the title "DE", is this a differential equation? Does your equation mean y''''+4y=0?

**ThePerfectHacker** · June 25th 2006, 04:38 PM

Originally Posted by Bert

Hello,

Given is $y^{4}+4y=0$ Who can I solove this? Who can do this step by step?

Greets Thanks .

It should it be
$y^{(4)}+4y=0$
---

If,
$y^{(4)}+4y=0$
The charachteristic equation is,
$k^4+4=0$
Thus,
$k^4=-4$
The solutions are,
$k=1+i,1-i,-1+i,-1-i$
Since the solutions are complex conjakates thus the general solution is,
$y=C_1e^x(\cos x+\sin x)+C_2e^x(\cos x-\sin x)$ $+C_3e^{-x}(\cos x+\sin x)+C_4e^{-x}(\cos x-\sin x)$

**Bert** · June 26th 2006, 12:15 AM

who do you get those complex solotions?

**Bert** · June 26th 2006, 02:18 AM

can you put $\frac{\sqrt{2}}{2}$ into the ct?

**ThePerfectHacker** · June 26th 2006, 10:36 AM

Originally Posted by Bert

who do you get those complex solotions?

But the rest you understand?
----------
Method 1) (Informal but it works!)
$k^4=-4$
Thus, (square root),
$k^2=\pm 2 i$

If $k^2=2i$
Take square roots,
$k=\pm \sqrt{2}\sqrt{i}$
But, $\sqrt{i}=\sqrt{2}/2\pm \sqrt{2}/2 i$ Thus,
$k=\pm \sqrt{2} (\sqrt{2}/2\pm\sqrt{2}/2 i)$
There are hence 4 possibilities (based on plus minus signs).
$k=-1-i,1+i,1-i,-1+i$

Finally if,
$k^2=-2i$ then by the same argument you get the same solutions, as before.

**ThePerfectHacker** · June 26th 2006, 10:48 AM

Method 2) (The is the elegant way. [My method if you are curious was to guess and play around with the roots cuz I was too lazy to work it out])

You have,
$k^4=-4$
Express this in polar form,
$k^4=4(\cos \pi+i\sin \pi)$
Now apply, de Moivre's theorem
And the roots are,
$4^{1/4} \left( \cos \left( \frac{\pi}{4}+\frac{2\pi k}{4}\right) +i\sin \left( \frac{\pi}{4}+\frac{2\pi k}{4} \right) \right)$ one for each value of $k=0,1,2,3$ . And $4^{1/4}=(4^{1/2})^{1/2}=2^{1/2}=\sqrt{2}$
You get your solutions.
$\sqrt{2} (\cos \pi/4+i\sin \pi/4)=1+i$
$\sqrt{2} (\cos 3\pi/4+i\sin 3\pi/4)=-1+i$
$\sqrt{2} (\cos 5\pi/4+i\sin 5\pi/4)=-1-i$
$\sqrt{2} (\cos 7\pi/4+i\sin 7\pi/4)=-1+i$

**Bert** · June 26th 2006, 12:54 PM

O yes thenk you very wel I see Greets.

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